题是这个吧?f(x)=√3sin2x+2cosx^2+mf(x)=√3sin2x+2cosx^2-1+1+m=√3sin2x-cos2x+1+m=2sin(2x-π/6)+1+mx∈[0,π/2]所以2x-π/6∈[0,5π/6]显然当2x-π/6=π/2取最大值即2sin(2x-π/6)=2所以f(x)max=2+1+m=3+m=6所...